Elegant Python code for Integer Partitioning

Question

I tried to write code to solve the standard Integer Partition problem (Wikipedia). The code I wrote was a mess. I need an elegant solution to solve the problem, because I want to improve my coding style. This is not a homework question.

Answer 1

A smaller and faster than Nolen's function:

def partitions(n, I=1):
    yield (n,)
    for i in range(I, n//2 + 1):
        for p in partitions(n-i, i):
            yield (i,) + p

Let's compare them:

In [10]: %timeit -n 10 r0 = nolen(20)
1.37 s ± 28.7 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [11]: %timeit -n 10 r1 = list(partitions(20))
979 µs ± 82.9 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [13]: sorted(map(sorted, r0)) == sorted(map(sorted, r1))
Out[14]: True

Looks like it's 1370 times faster for n = 20.

Anyway, it's still far from accel_asc:

def accel_asc(n):
    a = [0 for i in range(n + 1)]
    k = 1
    y = n - 1
    while k != 0:
        x = a[k - 1] + 1
        k -= 1
        while 2 * x <= y:
            a[k] = x
            y -= x
            k += 1
        l = k + 1
        while x <= y:
            a[k] = x
            a[l] = y
            yield a[:k + 2]
            x += 1
            y -= 1
        a[k] = x + y
        y = x + y - 1
        yield a[:k + 1]

Mine is not only slower, but requires much more memory (but apparently is much easier to remember):

In [18]: %timeit -n 5 r2 = list(accel_asc(50))
114 ms ± 1.04 ms per loop (mean ± std. dev. of 7 runs, 5 loops each)

In [19]: %timeit -n 5 r3 = list(partitions(50))
527 ms ± 8.86 ms per loop (mean ± std. dev. of 7 runs, 5 loops each)

In [24]: sorted(map(sorted, r2)) == sorted(map(sorted, r3))
Out[24]: True

---

You can find other versions on ActiveState: Generator For Integer Partitions (Python Recipe).

---

I use Python 3.6.1 and IPython 6.0.0.

Answer 2

While this answer is fine, I'd recommend skovorodkin's answer.

>>> def partition(number):
...     answer = set()
...     answer.add((number, ))
...     for x in range(1, number):
...         for y in partition(number - x):
...             answer.add(tuple(sorted((x, ) + y)))
...     return answer
... 
>>> partition(4)
set([(1, 3), (2, 2), (1, 1, 2), (1, 1, 1, 1), (4,)])

If you want all permutations(ie (1, 3) and (3, 1)) change answer.add(tuple(sorted((x, ) + y)) to answer.add((x, ) + y)

Answer 3

I've compared the solution with perfplot (a little project of mine for such purposes) and found that Nolen's top-voted answer is also the slowest.

Both answers supplied by skovorodkin are much faster. (Note the log-scale.)

---

To to generate the plot:

import perfplot
import collections


def nolen(number):
    answer = set()
    answer.add((number,))
    for x in range(1, number):
        for y in nolen(number - x):
            answer.add(tuple(sorted((x,) + y)))
    return answer


def skovorodkin(n):
    return set(skovorodkin_yield(n))


def skovorodkin_yield(n, I=1):
    yield (n,)
    for i in range(I, n // 2 + 1):
        for p in skovorodkin_yield(n - i, i):
            yield (i,) + p


def accel_asc(n):
    return set(accel_asc_yield(n))


def accel_asc_yield(n):
    a = [0 for i in range(n + 1)]
    k = 1
    y = n - 1
    while k != 0:
        x = a[k - 1] + 1
        k -= 1
        while 2 * x <= y:
            a[k] = x
            y -= x
            k += 1
        l = k + 1
        while x <= y:
            a[k] = x
            a[l] = y
            yield tuple(a[: k + 2])
            x += 1
            y -= 1
        a[k] = x + y
        y = x + y - 1
        yield tuple(a[: k + 1])


def mct(n):
    partitions_of = []
    partitions_of.append([()])
    partitions_of.append([(1,)])
    for num in range(2, n + 1):
        ptitions = set()
        for i in range(num):
            for partition in partitions_of[i]:
                ptitions.add(tuple(sorted((num - i,) + partition)))
        partitions_of.append(list(ptitions))
    return partitions_of[n]


perfplot.show(
    setup=lambda n: n,
    kernels=[nolen, mct, skovorodkin, accel_asc],
    n_range=range(1, 17),
    logy=True,
    # https://stackoverflow.com/a/7829388/353337
    equality_check=lambda a, b: collections.Counter(set(a))
    == collections.Counter(set(b)),
    xlabel="n",
)

Answer 4

I needed to solve a similar problem, namely the partition of an integer n into d nonnegative parts, with permutations. For this, there's a simple recursive solution (see here):

def partition(n, d, depth=0):
    if d == depth:
        return [[]]
    return [
        item + [i]
        for i in range(n+1)
        for item in partition(n-i, d, depth=depth+1)
        ]


# extend with n-sum(entries)
n = 5
d = 3
lst = [[n-sum(p)] + p for p in partition(n, d-1)]

print(lst)

Output:

[
    [5, 0, 0], [4, 1, 0], [3, 2, 0], [2, 3, 0], [1, 4, 0],
    [0, 5, 0], [4, 0, 1], [3, 1, 1], [2, 2, 1], [1, 3, 1],
    [0, 4, 1], [3, 0, 2], [2, 1, 2], [1, 2, 2], [0, 3, 2],
    [2, 0, 3], [1, 1, 3], [0, 2, 3], [1, 0, 4], [0, 1, 4],
    [0, 0, 5]
]

Answer 5

I'm a bit late to the game, but I can offer a contribution which might qualify as more elegant in a few senses:

def partitions(n, m = None):
  """Partition n with a maximum part size of m. Yield non-increasing
  lists in decreasing lexicographic order. The default for m is
  effectively n, so the second argument is not needed to create the
  generator unless you do want to limit part sizes.
  """
  if m is None or m >= n: yield [n]
  for f in range(n-1 if (m is None or m >= n) else m, 0, -1):
    for p in partitions(n-f, f): yield [f] + p

Only 3 lines of code. Yields them in lexicographic order. Optionally allows imposition of a maximum part size.

I also have a variation on the above for partitions with a given number of parts:

def sized_partitions(n, k, m = None):
  """Partition n into k parts with a max part of m.
  Yield non-increasing lists.  m not needed to create generator.
  """
  if k == 1:
    yield [n]
    return
  for f in range(n-k+1 if (m is None or m > n-k+1) else m, (n-1)//k, -1): 
    for p in sized_partitions(n-f, k-1, f): yield [f] + p

After composing the above, I ran across a solution I had created almost 5 years ago, but which I had forgotten about. Besides a maximum part size, this one offers the additional feature that you can impose a maximum length (as opposed to a specific length). FWIW:

def partitions(sum, max_val=100000, max_len=100000):
    """ generator of partitions of sum with limits on values and length """
    # Yields lists in decreasing lexicographical order. 
    # To get any length, omit 3rd arg.
    # To get all partitions, omit 2nd and 3rd args. 

    if sum <= max_val:       # Can start with a singleton.
        yield [sum]

    # Must have first*max_len >= sum; i.e. first >= sum/max_len.
    for first in range(min(sum-1, max_val), max(0, (sum-1)//max_len), -1):
        for p in partitions(sum-first, first, max_len-1):
            yield [first]+p

Answer 6

Much quicker than the accepted response and not bad looking, either. The accepted response does lots of the same work multiple times because it calculates the partitions for lower integers multiple times. For example, when n=22 the difference is 12.7 seconds against 0.0467 seconds.

def partitions_dp(n):
    partitions_of = []
    partitions_of.append([()])
    partitions_of.append([(1,)])
    for num in range(2, n+1):
        ptitions = set()
        for i in range(num):
            for partition in partitions_of[i]:
                ptitions.add(tuple(sorted((num - i, ) + partition)))
        partitions_of.append(list(ptitions))
    return partitions_of[n]

The code is essentially the same except we save the partitions of smaller integers so we don't have to calculate them again and again.

Answer 7

I think the recipe here may qualify as being elegant. It's lean (20 lines long), fast and based upon Kelleher and O'Sullivan's work which is referenced therein:

def aP(n):
    """Generate partitions of n as ordered lists in ascending
    lexicographical order.

    This highly efficient routine is based on the delightful
    work of Kelleher and O'Sullivan.

    Examples
    ========

    >>> for i in aP(6): i
    ...
    [1, 1, 1, 1, 1, 1]
    [1, 1, 1, 1, 2]
    [1, 1, 1, 3]
    [1, 1, 2, 2]
    [1, 1, 4]
    [1, 2, 3]
    [1, 5]
    [2, 2, 2]
    [2, 4]
    [3, 3]
    [6]

    >>> for i in aP(0): i
    ...
    []

    References
    ==========

    .. [1] Generating Integer Partitions, [online],
        Available: http://jeromekelleher.net/generating-integer-partitions.html
    .. [2] Jerome Kelleher and Barry O'Sullivan, "Generating All
        Partitions: A Comparison Of Two Encodings", [online],
        Available: http://arxiv.org/pdf/0909.2331v2.pdf

    """
    # The list `a`'s leading elements contain the partition in which
    # y is the biggest element and x is either the same as y or the
    # 2nd largest element; v and w are adjacent element indices
    # to which x and y are being assigned, respectively.
    a = [1]*n
    y = -1
    v = n
    while v > 0:
        v -= 1
        x = a[v] + 1
        while y >= 2 * x:
            a[v] = x
            y -= x
            v += 1
        w = v + 1
        while x <= y:
            a[v] = x
            a[w] = y
            yield a[:w + 1]
            x += 1
            y -= 1
        a[v] = x + y
        y = a[v] - 1
        yield a[:w]

Answer 8

Here is a recursive function, which uses a stack in which we store the numbers of the partitions in increasing order. It is fast enough and very intuitive.

# get the partitions of an integer

Stack = []
def Partitions(remainder, start_number = 1):
    if remainder == 0:
        print(" + ".join(Stack))
    else:
        for nb_to_add in range(start_number, remainder+1):
            Stack.append(str(nb_to_add))
            Partitions(remainder - nb_to_add, nb_to_add)
            Stack.pop()

When the stack is full (the sum of the elements of the stack then corresponds to the number we want the partitions), we print it, remove its last value and test the next possible value to be stored in the stack. When all the next values have been tested, we pop the last value of the stack again and we go back to the last calling function. Here is an example of the output (with 8):

Partitions(8)
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1 + 1 + 2
1 + 1 + 1 + 1 + 1 + 3
1 + 1 + 1 + 1 + 2 + 2
1 + 1 + 1 + 1 + 4
1 + 1 + 1 + 2 + 3
1 + 1 + 1 + 5
1 + 1 + 2 + 2 + 2
1 + 1 + 2 + 4
1 + 1 + 3 + 3
1 + 1 + 6
1 + 2 + 2 + 3
1 + 2 + 5
1 + 3 + 4
1 + 7
2 + 2 + 2 + 2
2 + 2 + 4
2 + 3 + 3
2 + 6
3 + 5
4 + 4
8

The structure of the recursive function is easy to understand and is illustrated below (for the integer 31):

remainder corresponds to the value of the remaining number we want a partition (31 and 21 in the example above). start_number corresponds to the first number of the partition, its default value is one (1 and 5 in the example above).

If we wanted to return the result in a list and get the number of partitions, we could do this:

def Partitions2_main(nb):
    global counter, PartitionList, Stack
    counter, PartitionList, Stack = 0, [], []
    Partitions2(nb)
    return PartitionList, counter

def Partitions2(remainder, start_number = 1):
    global counter, PartitionList, Stack
    if remainder == 0:
        PartitionList.append(list(Stack))
        counter += 1
    else:
        for nb_to_add in range(start_number, remainder+1):
            Stack.append(nb_to_add)
            Partitions2(remainder - nb_to_add, nb_to_add)
            Stack.pop()

Last, a big advantage of the function Partitions shown above is that it adapts very easily to find all the compositions of a natural number (two compositions can have the same set of numbers, but the order differs in this case): we just have to drop the variable start_number and set it to 1 in the for loop.

# get the compositions of an integer

Stack = []
def Compositions(remainder):
    if remainder == 0:
        print(" + ".join(Stack))
    else:
        for nb_to_add in range(1, remainder+1):
            Stack.append(str(nb_to_add))
            Compositions(remainder - nb_to_add)
            Stack.pop()

Example of output:

Compositions(4)
1 + 1 + 1 + 1
1 + 1 + 2
1 + 2 + 1
1 + 3
2 + 1 + 1
2 + 2
3 + 1
4

Answer 9

# -*- coding: utf-8 -*-
import timeit

ncache = 0
cache = {}


def partition(number):
    global cache, ncache
    answer = {(number,), }
    if number in cache:
        ncache += 1
        return cache[number]
    if number == 1:
        cache[number] = answer
        return answer
    for x in range(1, number):
        for y in partition(number - x):
            answer.add(tuple(sorted((x, ) + y)))
    cache[number] = answer
    return answer


print('To 5:')
for r in sorted(partition(5))[::-1]:
    print('\t' + ' + '.join(str(i) for i in r))

print(
    'Time: {}\nCache used:{}'.format(
        timeit.timeit(
            "print('To 30: {} possibilities'.format(len(partition(30))))",
            setup="from __main__ import partition",
            number=1
        ), ncache
    )
)

or https://gist.github.com/sxslex/dd15b13b28c40e695f1e227a200d1646

Answer 10

I don't know if my code is the most elegant, but I've had to solve this many times for research purposes. If you modify the

sub_nums

variable you can restrict what numbers are used in the partition.

def make_partitions(number):
    out = []
    tmp = []
    sub_nums = range(1,number+1)
    for num in sub_nums:
        if num<=number:
            tmp.append([num])
        for elm in tmp:
            sum_elm = sum(elm)
            if sum_elm == number:
                out.append(elm)
            else:
                for num in sub_nums:
                    if sum_elm + num <= number:
                         L = [i for i in elm]
                         L.append(num)
                         tmp.append(L)
    return out

Answer 11

F(x,n) = \union_(i>=n) { {i}U g| g in F(x-i,i) }

Just implement this recursion. F(x,n) is the set of all sets that sum to x and their elements are greater than or equal to n.